Without calculation, find one eigenvalue and two linearly independent eigenvectors of $A = \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$ Justify your answer.

One eigenvalue of A is λ = 0 because the columns of A are linearly dependent.

Two linearly independent eigenvectors of A are

Question :

Without calculation, find one eigenvalue and two linearly independent eigenvectors of $a = \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}$ justify your answer.

one eigenvalue of a is λ = 0 because the columns of a are linearly dependent.

two linearly independent eigenvectors of a are

Without calculation, find one eigenvalue and two linearly independent eigenvecto | Doubtlet.com

Solution:

Neetesh Kumar | October 16, 2024

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This is the solution to Math2B Course: Linear Algebra
Assignment: Ch5 Section 1 Question Number 8
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Step-by-step solution:

Eigenvalue and Eigenvector calculator

The matrix $A$ is symmetric, and all the matrix's columns are identical, indicating that the columns of $A$ are linearly dependent.

Eigenvalue:

Since the columns of $A$ are linearly dependent, one of the eigenvalues of $A$ is $\lambda = 0$ .

Eigenvectors:

To find two linearly independent eigenvectors, we observe that any vector that sums the entries to zero will be an eigenvector associated with $\lambda = 0$ . Two such vectors are:

$\mathbf{v_1} = \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \mathbf{v_2} = \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}$

These vectors are linearly independent because they are not scalar multiples of each other, and the sum of their components is zero.

Justification:

Eigenvalue: $\lambda = 0$ because the columns of $A$ are linearly dependent.
Eigenvectors: The entries of each vector sum to zero make them valid eigenvectors for $\lambda = 0$ .

Final Answer:

One eigenvalue of $A$ is: $\boxed{0}$
Two linearly independent eigenvectors of $A$ are: $\boxed{\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}}$

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